Let us assume that each droplet of ink is a sphere of diameter D
as it flies
towards the paper, then creates a spherical dot about 20 percent
larger than D when it strikes the paper.
How large must the drop be to yield a resolution of 300 dots per inch?

The printer shoots droplets horizontally towards the paper.
When aiming for a particular spot on the paper,
is it necessary to take into account the deflection of the
drops due to gravity during the flight?
Well, how far does the droplet fall downwards during its flight?

Now, since the droplet itself has a radius of R = 35 microns,
this deflection is only a tiny fraction of the drop's size.
We can ignore the effect of gravity on the drop without incurring
any perceptible change in the placement of dots on the paper.

How many drops must the ink jet produce each second?
It depends on how fast the printer must create finished pages.
Suppose that the goal is one page in 60 seconds.
The most difficult page to print is one which is covered completely
in black ink.
Such a page would require

In order to print a page more rapidly, one might mount a number of
ink guns on a single head. Some ink jet printers have 30 guns
per head; they might be able to print a single page in two seconds.

How much space lies between each of the droplets as they fly through
the air towards the paper?
Each droplet takes t = 0.00015 seconds to cover the gap
between head and paper. During that time, there must be

How much ink does the printer use?
The simplest, but most wasteful, method of operation is to
produce ink droplets continuously during the printing process.
Suppose that a printer does so -- how much ink does it consume
per second?

How long does a typical ink-jet cartridge last?
Let's do a very rough, order-of-magnitude estimate.
Suppose a printer is active for two minutes per hour,
8 hours a day, 5 days a week.
Then each week, it runs for 80 minutes. During that time, it
uses a total of

Real ink jets (probably) have intelligent algorithms which
turn off the ink gun when it's not needed for a while
(between words, or between characters).
They are more efficient than the simple-minded printer
in this example.

In order to control the motion of ink droplets,
they must be given an electric charge.
The electron gun can shoot electrons at a droplet
as it moves past the gun in order to give it
a negative electric charge.
How much current does it take?

The droplet has a diameter D = 70 microns,
and moves with a speed v = 20 m/s past the electron gun.
Electrons fired straight out of the gun will hit the
droplet during a window of time

The electron gun must give the droplet a charge of about
q = 1.9 x 10^(-10) Coulombs
in order to permit the capacitor to deflect it into the gutter.
Assuming that all electrons which strike the droplet are absorbed
into it,
the current of electron beam must therefore be

Of course, some of the droplets must not be charged.
The electron gun must be able to switch on and off at precise
intervals.
Specifically, the gun must turn on and off in the amount of
time between one drop and the next.
The space between drops is about 143 microns, so the time interval
between drops -- during which the electron gun must be able to
switch on/off -- is

The ink gun sends a steady stream of droplets towards the
paper.
If they all reached the paper, they would leave a solid black area.
In order to form letters, numbers, and images, the printer
must somehow prevent some of the droplets from reaching
the paper; only a small fraction of all the droplets strike
the paper for a typical document.

The capacitor performs this task:
it creates a uniform electric field between its
plates.
Neutral droplets pass through the field unaffected and strike the paper.
Droplets with a negative charge, however, are deflected downwards
(towards the positive plate) and fall into a reservoir called
the gutter.

How strong must the electric field be in order to deflect
charged droplets strongly enough to force them into the gutter?
We know

If a droplet has a velocity directed downwards at a 45 degree angle
when it leaves the capacitor, then it will move downwards by 0.5 mm
(the distance from the original path to the gutter) in the same
time that it moves forwards by 0.5 mm (the distance from the
right edge of the capacitor to the end of the gutter).